If 150mL of .055M copper(II) chloride solution reacts, how many mL of .100M silver nitrate is needed to exactly react with it?We went over this topic in class today but I just don't understand how to do this problem
So first, since you have the volume and molarity for the copper chloride solute, you find the moles for copper chloride. I will be using n to symbolize moles. Remember that the equation for Molarity is Molarity=number of moles of solute divided by the Volume of solution. So your first step would be solving for moles of the copper chloride. That would look like the following: .055M= n/150mL of copper chloride By multiplying 150mL times .055M, you would get 8.25moles. Next, you plug the moles into the Molarity equation again, except now you are solving for the volume of silver nitrate. That would look like the following: .100M=8.25moles divided by the Volume of silver nitrate You solve for volume by dividing 8.25 by .100M to get 82.5. This means that it takes 82.5 mL of silver nitrate to react with the copper chloride. Hope this helps!
